(x+2⼀x+1)-(x+4⼀x+3)-(x+3⼀x+2)+(x+5⼀x+4)

初二下的分解因式。可能要用到简便运算。可能啊!
2024-12-16 14:08:36
推荐回答(1个)
回答1:

(x+2/x+1)-(x+4/x+3)-(x+3/x+2)+(x+5/x+4)
=[(x+1/x+1)+(1/x+1)]- [(x+3/x+3)+(1/x+3)]- [(x+2/x+2)+(1/x+2)]+ [(x+4/x+4)+(1/x+4)]
=(1/x+1)- (1/x+3) - (1/x+2) +(1/x+4)
=[(1/x+1)+(1/x+4)]- [(1/x+2)+(1/x+3)]
=[(x+4+x+1)/[(x+1)(x+4)] [(x+3+x+2)/(x+2)(x+3)]
=(2x+5){ [1/(x+1)(x+4)] -[1/(x+2)(x+3)]}
这里把分母整理 演算 得
=(2x+5) [2/(x+1)(x+2)(X+3)(x+4)]
=(4x+10)/(x+1)(x+2)(X+3)(x+4)